Theorem: Scaled turnover with recovery and a one-qubit cutoff

Let

Choose $\zeta >0$ and

Declare

A set of Haar devices of measure at least $1/(4d)$ satisfies:

1. $z_k(U)\le \zeta$ for every $k_{\ast }\le k\le K$;

2. one fixed decoder core has half-diamond error at most $\min \{a_d,\sqrt{2\zeta }\}$ at every retained cut;

3. for the fixed pure input,

   $$S({P_{k_{\mathrm{peak}}}}^U)>{\log }_2(h\sqrt{d}/2)\ge \max \{S({P_{k_{\ast }}}^U),S({P_{k_{+}}}^U),S({P_K}^U)\};$$

4. $B_K$ is one physical qubit.

The ensemble mean has the same strict peak relative to $k_{\ast }$ and $k_{+}$.

Proof. Write $t=\sqrt{d}=2^s$. The first three cuts have dimensions

The cutoff has $b_K=2$.

At the first cut, (eq:momentbound) gives

Therefore

At the peak, put $\mu =ht/2$. Equation (eq:purity) yields

Hence

Markov's inequality implies

On this event, $S({P_{\mathrm{peak}}}^U)>{\log }_2\mu$. The rank bound at $k_{\ast }$ and $k_{+}$ is $\mu$; at $K$ it is $2h\le \mu$. Subtracting the possible leakage-failure set leaves measure greater than $1/(4d)$.

Nested monotonicity propagates leakage, and Proposition prop:decoder supplies the common receiver policy. Also $p_{\mathrm{q}}<1/\mu$, so the mean entropy lower bound is strictly above ${\log }_2\mu$ at the peak.